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jeka94
3 years ago
11

1.8 = 2.1h - 5.7 - 4.6h h= __

Mathematics
1 answer:
Blababa [14]3 years ago
7 0

Answer:

how are you doing

Step-by-step explanation:

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Simplify 12/5(10/3b+15/4) .
Elodia [21]
Step 1. Simplify 10/3b to 10b/3
12/5(10b/3 + 15/4)
Step 2. Distribute 
8b + 9

7 0
3 years ago
Read 2 more answers
How do I find the average rate of change?​
Alenkasestr [34]

Answer:

add them all up together and then divide by the number of items

Step-by-step explanation:

8 0
3 years ago
13. Suppose the probability of passing a driving test is 65%. (a) On a certain afternoon 15 people are tested. (i) What is the e
Alex_Xolod [135]

Answer: 9

Step-by-step explanation:

Based on the information given in the question, the expected number of people who pass will be calculated as:

= Probability of passing the driving test × Number of people tested

= 65% × 15

= 0.65 × 15

= 9.75

Therefore, expected number of people who pass is 9.

8 0
3 years ago
UNA POBLACION DE CIERTA CIUDAD EN EL AÑO 2000 ERA DE 250,000 HABITANTES CON UNA TASA DE CRECIMIENTO RELATIVO DEL 2% , DETERMINA
Fofino [41]

Answer:

La población que se espera para el 2012, es 317,060

Step-by-step explanation:

Si la población tiene un crecimiento relativo del 2%, entonces cada año, la población es un 2% más grande que el año anterior.

Definamos P(t) = población después de t años

Entonces, si al año t = 0 (que corresponde con el año 2000) la población es A

P(0) = A

Un año después, en t = 1, la población incrementa en un 2%

P(1) = A + (2%/100%)*A = A + (0.02)*A = A*(1.02)

Otro año después, en t = 2, la población incrementa un 2% de vuelta.

P(2) = A*(1.02) + (2%/100%)*A(1.02) = A*(1.02) + 0.02*A*(1.02)

      = A*(1.02)*(1.02) = A*(1.02)^2

Ya podemos notar un patrón, la población en el año t va a ser:

P(t) = A*(1.02)^t

Sabemos que en t = 0 (el año 2000) la población es 250,000

Entonces A = 250,000

P(t) = 250,000*(1.02)^t

Ahora queremos calcular la población en el año 2012

entonces si t = 0 es el 2000

2012 esta representado con t = 12

Reemplazando eso en la ecuación, obtenemos:

P(12) = 250,000*(1.02)^12 = 317,060.4

Como esto es una población tenemos que redondearlo al próximo número entero, como el primer digito después del punto es 4, redondeamos para abajo.

Entonces la población que se espera para el 2012 es: 317,060

5 0
3 years ago
Suppose that in one region of the country the mean amount of credit card debt perhousehold in households having credit card debt
kvv77 [185]

Answer:

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 15250, \sigma = 7125, n = 1600, s = \frac{7125}{\sqrt{1600}} = 178.125

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly

This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So

X = 1900

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1900 - 1600}{178.125}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535.

X = 1300

Z = \frac{X - \mu}{s}

Z = \frac{1300 - 1600}{178.125}

Z = -1.68

Z = -1.68 has a pvalue of 0.0465.

0.9535 - 0.0465 = 0.907.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

7 0
3 years ago
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